Chemistry, 10.07.2019 09:00 chris199825

Compound a, c8h10, undergoes nitration to give 2 products. the 1h nmr spectrum of a shows a complex five-proton multiplet at 7.27-7.40 δ, a two-proton quartet at 2.60 δ, and a three-proton triplet at 1.25 δ. what is the structure of a?

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Answer from: eddsworldfrantic

Compound A is Ethylbenzene.

Explanation:

The structure of Ethylbenzene was elucidated and finalized based on following facts:

1) The presence of multiplet signal ranging from δ 7.27 to 7.40 ppm confirms the presence of aromatic protons. Also the statement says that this signal is for 5 protons indicating that the benzene ring is mono substituted.

2) The substituted group on benzene is Ethyl because the quartet for two protons shows that two protons (-CH₂-) are coupled with a CH₃ group. Hence, the ethyl group gives one quartet and one triplet signal.

3) The chemical reaction stated in statement is Nitration reaction. Also, the alkyl group (ethyl) is ortho-para directing in nature. Hence, it will give two products i.e. 1-Ethyl-2-nitrobenzene and 1-Ethyl-4-nitrobenzene respectively.

Attached figure gives the calculated NMR spectrum of Ethylbenzene and the chemical shifts calculated are almost near to the given values. Also, the two products formed are shown.

Compound a, c8h10, undergoes nitration to give 2 products. the 1h nmr spectrum of a shows a complex

Answer from: Quest

the major pests we deal with are insects, weeds and fungal diseases. a pesticide is a product, whether organic or chemical, used to kill the target pest. there are other techniques gardeners should use that can prevent the severe infestations that make the use of pesticides necessary

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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